Помогите решить неопределенные интегралы: 1)

Решите задачу:

$1)\;\int\frac{dx}{\sqrt{6x-x^2}}=\int\frac{dx}{\sqrt{9-9+6x-x^2}}=\int\frac{dx}{\sqrt{9-(x-3)^2}}=\left(\begin{array}{c}u=x-3\\du=dx\end{array}\right)=\\=\int\frac{du}{\sqrt{9-u^2}}=\int\frac{du}{\sqrt{3^2-u^2}}=\arcsin\frac{u}3+C=\arcsin\left(\frac{x-3}3\right)+C\\\\2)\;\int\cos^54x\;dx=\left(\begin{array}{c}u=4x\\du=4dx\end{array}\right)=\frac14\int\cos^54x\;dx\\\int\cos^nax\;dx=\frac{\cos^{n-1}ax\cdot\sin ax}{an}+\frac{n-1}n\int\cos^{n-2}ax\;dx$

$\frac14\int\cos^54x\;dx=\frac14\left(\frac15\sin u\cos^4u+\frac45\int\cos^3u\;du\right)=\\=\frac1{20}\sin u\cos^4u+\frac15\int\cos^3u\;du=\frac1{20}\sin u\cos^4u+\frac1{15}\sin u\cos^2u+\\+\frac2{15}\int\cos u\;du=\frac1{20}\sin u\cos^4u+\frac1{15}\sin u\cos^2u+\frac2{15}\sin u+C=\\=\frac1{960}\left(150\sin4x+25\sin12x+3\sin20x\right)+C\\\\3)\;\int\frac{x\;dx}{x^2+10}=\left(\begin{array}{c}u=x^2+10\\du=2x\;dx\end{array}\right)=\frac12\int\frac{du}u=\frac12\ln u+C=\\=\frac12\ln(x^2+10)+C$

$4)\;\int\frac{dx}{2x+3}=\left(\begin{array}{c}u=2x+3\\du=2dx\end{array}\right)=\frac12\int\frac{du}u=\frac12\ln u+C=\frac12\ln(2x+3)+C$